In a series of three trials, the probability of exactly two successes in nine times is as large as the probability of three successes. Then, the probability of success in each trial is :

Question

In a series of three trials, the probability of exactly two successes in nine times is as large as the probability of three successes. Then, the probability of success in each trial is : (A)  (1/2)  (B)  (1/3)  (C)  (1/4)  (D)  (3/4)

Tardigrade Admin · Accepted Answer

Let x be the probability of success in each trial, then (1 - x) will be the probability of failure in each trial. Thus, probability of exactly two successes in a series of trials = P( barE1 E2 E3+E1 barE2 E3+E1 E2 barE3) =(1-x) x ⋅ x+x(1-x) x+x ⋅ x(1-x) =3 x2(1-x) and the probability of three successes P(E1 E2 E3)=x ⋅ x ⋅ x=x3 According to question ⇒ 9 x3 =3 x2(1-x) 3 x =1-x ⇒ 4 x=1 ⇒ x=(1/4) Hence, the probability of success in each trial is (1/4)

Q. In a series of three trials, the probability of exactly two successes in nine times is as large as the probability of three successes. Then, the probability of success in each trial is :

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{3}{4}$

Q. In a series of three trials, the probability of exactly two successes in nine times is as large as the probability of three successes. Then, the probability of success in each trial is :

21​

31​

41​

43​

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

$\frac{3}{4}$